How do you solve Classes With at Least 5 Students on LeetCode?

Classes With at Least 5 Students (LeetCode #596) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using GROUP BY and HAVING in SQL can significantly reduce the complexity of counting and filtering operations.

What is the time complexity of Classes With at Least 5 Students?

The optimal solution for Classes With at Least 5 Students runs in O(n) time and uses O(n) space. The time complexity is O(n) because we only need to scan the data once to group and count students, which is much more efficient.

What is the brute force solution for Classes With at Least 5 Students?

The brute force approach for Classes With at Least 5 Students is Brute Force, with O(n²) time complexity and O(1) space complexity. In this approach, we will check each class and count the number of students enrolled in it. If a class has at least five students, we will include it in our result. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Classes With at Least 5 Students?

Classes With at Least 5 Students uses the following concepts: Database. The recommended patterns to study are: Hash Map, Aggregation.

What are the interview tips for Classes With at Least 5 Students?

Always think about how to aggregate data efficiently before diving into complex logic. Practice writing SQL queries to become comfortable with GROUP BY and HAVING clauses. Explain your thought process clearly during the interview; it shows your understanding of the problem.

What are common mistakes when solving Classes With at Least 5 Students?

Not using GROUP BY correctly, leading to incorrect counts. Forgetting to use HAVING to filter aggregated results.

#596

Classes With at Least 5 Students

Easy

DatabaseHash MapAggregation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Using SQL's GROUP BY and HAVING clauses allows us to efficiently group students by class and filter those with at least five students in a single query. This is more efficient than checking each class individually.

⚙️

Algorithm

3 steps

1Step 1: Use GROUP BY to group records by class.
2Step 2: Use COUNT to count the number of students in each class.
3Step 3: Use HAVING to filter groups with a count of 5 or more.

solution.py1 lines

1SELECT class FROM Courses GROUP BY class HAVING COUNT(student) >= 5;

ℹ

Complexity note: The time complexity is O(n) because we only need to scan the data once to group and count students, which is much more efficient.

1Using GROUP BY and HAVING in SQL can significantly reduce the complexity of counting and filtering operations.
2Understanding how to aggregate data is crucial for efficiently solving problems involving counts.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.